Context: these notes, page 7, last paragraph (between equations 2.7 and 2.8).
Given the $\mathcal{C}^r$ function with compact support $\varphi:\mathbb{R}^d\to \mathbb{R}$, we define $$\varphi_x^n:=2^{nd/2}\varphi(2^n(\cdot -x)).$$
Next, for some $\mathcal{C}^r$ function with compact support $\psi$, we try to bound
$$|\langle\varphi_x^n,\psi\rangle|.$$
Substituting in the definition of $\varphi_x^n$, we get
$$|\langle\varphi(2^n(\cdot -x)),\psi\rangle|$$ $$= 2^{nd/2}|\int \varphi(2^n(\cdot -x))\psi|$$ Since both $\varphi$ and $\psi$ are integrable and with compact support, I would expect the integral to be bounded. Like this, I would obtain
$$|\langle\varphi_x^n,\psi\rangle| \lesssim 2^{nd/2}.$$
However, the paper gives
$$|\langle\varphi_x^n,\psi\rangle| \lesssim 2^{-nd/2}.$$
Why might that be?
Notice that the paper is a far better majoration. A change of variable will help us to get it.
$$\int_{\Bbb{R}^d } \varphi(2^n(y-x)) \psi(y) dy = \int_{\Bbb{R}^d } \varphi(y-2^nx) \psi(\frac{y}{2^n}) (2^{-n})^d dy$$
Now $\psi$ is bounded, and $\varphi$ is bounded with compact support $K$, so
$$ \left| \int_{\Bbb{R}^d } \varphi(2^n(y-x)) \psi(y) dy \right| = \left| \int_{\Bbb{R}^d } \varphi(y-2^nx) \psi(2^{-n}y) 2^{-nd} dy \right|$$
$$= \left| \int_{K+2^nx } \varphi(y-2^nx) \psi(2^{-n}y) 2^{-nd} dy \right|$$
$$\leq 2^{-nd} \int_{K+2^nx} \|\varphi\|_\infty \| \psi \|_\infty dy$$
$$\leq 2^{-nd} \mu(K) \|\varphi\|_\infty \| \psi \|_\infty$$
And if you multiply this by $2^{\frac{nd}{2}}$, you get the given coefficient